1. **State the problem:** Solve for $2^x$ given the equation $4^x = 32^{x-3}$. 2. **Rewrite bases as powers of 2:** - $4 = 2^2$ so $4^x = (2^2)^x = 2^{2x}$. - $32 = 2^5$ so $32^{x-3} = (2^5)^{x-3} = 2^{5(x-3)}$. 3. **Set the exponents equal since bases are the same:** $$2x = 5(x-3)$$ 4. **Solve for $x$:** $$2x = 5x - 15$$ $$2x - 5x = -15$$ $$\cancel{2x} - \cancel{5x} = -15$$ $$-3x = -15$$ $$x = \frac{-15}{-3} = 5$$ 5. **Find $2^x$:** $$2^x = 2^5 = 32$$ **Final answer:** $2^x = 32$ which corresponds to option D.